Question

Which expression is a cube root of -1 + i sqrt(3)?

A. cubert(2) (cos(120degrees) + i sin(120 degrees))
B. cubert(2) (cos(40 degrees) + i sin(40 degrees))
C. cubert(2) (cos(280 degrees) + i sin(280 degrees))
D. cubert(2) (cos(320 degrees) + i sin(320 degrees))

***My Answer: D***
Please Help!!!

Answer 1

\[-1+\sqrt{3}=r \left( \cos \theta+\iota \sin \theta \right)\]
\[r \cos \theta =-1,r \sin \theta =\sqrt{3}\]
square and add and find r
divide find theta

Answer 2

So I would do r = -1-cos(theta), r = 3sin^2(theta)?

Answer 3

\[r^2\cos ^2\theta =1,r^2\sin ^2\theta=3\]
\[r^2\left( \cos ^2\theta+\sin ^2\theta \right)=1+3=4\]
r=2

Answer 4

Oh ok

Answer 5

r is always positive ,so \[\cos~ \theta~ is~ negative~and~\sin \theta~is~positive.\]
so angle lies in second quadrant.
now divide to find theta

Answer 6

Would \[\theta = 120\] be the angle?

Answer 7

no guess ,show me your calculations.

Answer 8

Ok

\[r ^{2}(\cos ^{2}\theta + \sin ^{2}\theta)\]\[2 (\cos ^{2}\theta + \sin ^{2}\theta)\]

\[-1 + i \sqrt{3}\]

\[\theta = \frac{ \sqrt{3} }{ -1}\]\[\theta = -\frac{ \pi }{ 3}\]

Answer 9

That is all I know of. Not exactly sure if it is correct. Sorry

Answer 10

\[\frac{ r \sin \theta }{ r \cos \theta }=\frac{ \sqrt{3} }{ -1 }=-\sqrt{3}=-\tan 60=\tan \left( 180-60\right) =\tan 120\]

Answer 11

\[\tan \theta=\tan 120,\theta=120\]

Answer 12

Oh wow I was far off but at the same time I kinda sorta knew what I was doing. Haha. I answer would now be A I'm assuming. Thank you for taking the time to explain this to me. I desperately needed the help.

Answer 13

\[-1+i \sqrt{3}=2\left( \cos 120+i \sin 120 \right)\]

Answer 14

yw

Answer 15

Oh wait the answer would be B

Answer 16

oh sorry you wanted cube root ,i have not noticed

Answer 17

It's ok :) I already hit the submit button haha

Answer 18

Wouldn't you just divide 120/3 to get 40?

Answer 19

\[1+i \sqrt{3}=2\left( \cos( 120+360n)+i \sin \left( (120+360n \right) \right)\]
\[or Z ^{\frac{ 1 }{ 3 }}=2^1/3e ^{\frac{ 360n+120 }{ 3 }}\]
put n=0,1,2 ,you get all the three cube roots

Answer 20

\[\left( 1+i \sqrt{3} \right)^{\frac{ 1 }{ 3 }}=2^{\frac{ 1 }{ 3 }}\left\{ \cos \left( \frac{ 360n+120 }{ 3 }+i \sin \left( \frac{ 360n+120 }{ 3 } \right)\right)\right\}\]
put n=0,1,2 you get all the values.

Answer 21

Thank you so much! I very much appreciate it!

Answer 22

yw